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two questions about vector fields

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Phil

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Apr 27, 2008, 12:14:49 AM4/27/08
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1.
Consider two complete vector fields on R^n. Is it known if their Lie
bracket is complete?


2.
On a (finite-dimensional) connected manifold, is it known when the Lie
algebra of all vector fields that commute with a fixed vector field is
finite-dimensional?

More generally, is it known when the Lie algebra of all vector fields
V that preserve a given tensor X, in the sense that the Lie derivative
L_V X = 0, is finite-dimensional?

José Carlos Santos

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Apr 30, 2008, 6:01:27 PM4/30/08
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On 27-04-2008 5:14, Phil wrote:

> On a (finite-dimensional) connected manifold, is it known when the Lie
> algebra of all vector fields that commute with a fixed vector field is
> finite-dimensional?

Yes, it is known (or, it least, I know it) that it is not
finite-dimensional. Just take the null vector field; all vector fields
commute with it, but they do not form a finite-dimensional space.

If this example is too silly for your taste, then consider the vector
field d/dx in R^2. Then all the vector fields of the form f(y)d/dy
commute with it, but they also do not form a finite-dimensional space.

Best regards,

Jose Carlos Santos

Maarten Bergvelt

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Apr 30, 2008, 9:30:11 PM4/30/08
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In article <fv4vql$9j7$1...@news.ks.uiuc.edu>, Phil wrote:
> More generally, is it known when the Lie algebra of all vector fields
> V that preserve a given tensor X, in the sense that the Lie derivative
> L_V X = 0, is finite-dimensional?

The Lie group of symplectomorphisms of a cotangent bundle is infinite
dimensional, (it contains the diffeomorphism group of the base) so
this gives an infinite Lie algebra of vector fields preserving the
symplectic structure. (This should be in Arnold, or
Marsden-Weinstein). On the other hand the group of symmetries of a
Riemannian manifold is finite dimensional when not trivial. This is
discussed in books on general relativity, maybe Misner, Thorne,
Wheeler? Also Helgason (Diff. Geom., Lie Groups,etc) might have a
discussion of this.

--
Maarten Bergvelt

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